If \(\displaystyle{\lim_{x \to \infty}}\)(\({x^3+1\over x^2+1}-(ax+b)\)) = 2, then find the value of a and b.

Solution :

\(\displaystyle{\lim_{x \to \infty}}\)(\({x^3+1\over x^2+1}-(ax+b)\)) = 2

\(\implies\) \(\displaystyle{\lim_{x \to \infty}}\)\(x^3(1-a)-bx^2-ax+(1-b)\over x^2+1\) = 2

\(\implies\) \(\displaystyle{\lim_{x \to \infty}}\)\(x(1-a)-b-{a\over x}+{(1-b)\over x^2}\over 1+{1\over x^2}\) = 2

\(\implies\) 1 – a = 0, -b = 2 \(\implies\) a = 1, b = -2


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